Algebraic Fractions, Rational Equations, and Mathematical Proofs

Factorisation and Simplification of Algebraic Fractions

• Fundamental Principles of Algebraic Fraction Simplification:

  • An algebraic fraction (or rational expression) represents the quotient of two polynomials, expressed in the form $\frac{P(x)}{Q(x)}$, where the denominator $Q(x) \neq 0$.

• Simplification requires factorising both the numerator and the denominator completely into their prime polynomial factors and dividing out all common factors (utilising the identity $\frac{a \cdot k}{b \cdot k} = \frac{a}{b}$, provided $k \neq 0$).

  • Values of the variable that cause any denominator to equal zero are strictly excluded from the domain of the expression.

• Standard Factorisation Techniques Applied:

  • Common Monomial Factorisation: Extracting the greatest common factor (GCF) from terms, following the distributive identity $ax + ay = a(x + y)$.
  • Difference of Two Squares: Applying the identity $a^2 - b^2 = (a - b)(a + b)$.
  • Quadratic Trinomial Factorisation: Factoring monic quadratics of the form $x^2 + Bx + C$ into $(x + p)(x + q)$, where $p + q = B$ and $p \times q = C$.

• Worked Simplification Problems:

  • Expression: $\frac{x + 5}{x^2 - 25}$
  • The denominator represents a difference of two squares: $x^2 - 25 = (x - 5)(x + 5)$.
  • Rewriting the expression with factorised denominator: $\frac{x + 5}{(x - 5)(x + 5)}$.
  • Dividing out the common factor $(x + 5)$: $\frac{1}{x - 5}$, subject to domain restrictions $x \neq 5$ and $x \neq -5$.

• Expression: $\frac{2x - 10}{x^2 - 5x}$

  • Factoring the greatest common factor $2$ from the numerator: $2(x - 5)$.
  • Factoring the greatest common factor $x$ from the denominator: $x(x - 5)$.
  • Rewriting: $\frac{2(x - 5)}{x(x - 5)}$.
  • Cancelling the shared binomial factor $(x - 5)$: $\frac{2}{x}$, where $x \neq 0$ and $x \neq 5$.

• Expression: $\frac{2x^2 - 8}{x + 2}$

  • Factoring the common numerical factor $2$ from the numerator: $2(x^2 - 4)$.
  • Factoring the resulting difference of squares: $2(x - 2)(x + 2)$.
  • Rewriting: $\frac{2(x - 2)(x + 2)}{x + 2}$.
  • Cancelling the common factor $(x + 2)$: $2(x - 2) = 2x - 4$, where $x \neq -2$.

• Expression: $\frac{3x - 12}{(x - 4)^2}$

  • Factoring the numerator: $3(x - 4)$.
  • The denominator is an expanded repeated binomial factor: $(x - 4)(x - 4)$.
  • Rewriting: $\frac{3(x - 4)}{(x - 4)^2}$.
  • Cancelling one power of $(x - 4)$: $\frac{3}{x - 4}$, where $x \neq 4$.

• Expression: $\frac{x^2 - 4x + 3}{x^2 - 3x}$

  • Factoring the monic quadratic in the numerator with factors of $+3$ that sum to $-4$ ($-1$ and $-3$): $(x - 1)(x - 3)$.
  • Factoring $x$ from the denominator: $x(x - 3)$.
  • Rewriting: $\frac{(x - 1)(x - 3)}{x(x - 3)}$.
  • Cancelling the common factor $(x - 3)$: $\frac{x - 1}{x}$, where $x \neq 0$ and $x \neq 3$.

• Expression: $\frac{x^2 - 2x - 8}{x^2 + 5x + 6}$

  • Factoring the numerator with factors of $-8$ summing to $-2$ ($-4$ and $+2$): $(x - 4)(x + 2)$.
  • Factoring the denominator with factors of $+6$ summing to $+5$ ($+2$ and $+3$): $(x + 2)(x + 3)$.
  • Rewriting: $\frac{(x - 4)(x + 2)}{(x + 2)(x + 3)}$.
  • Cancelling the common factor $(x + 2)$: $\frac{x - 4}{x + 3}$, where $x \neq -2$ and $x \neq -3$.

Multiplication and Division of Algebraic Fractions

• Rules of Operations on Rational Fractions:

  • Multiplication: Multiply numerators directly and denominators directly: $\frac{A}{B} \times \frac{C}{D} = \frac{A \cdot C}{B \cdot D}$.
  • Division: Convert the operation into multiplication by the reciprocal of the divisor: $\frac{A}{B} \div \frac{C}{D} = \frac{A}{B} \times \frac{D}{C} = \frac{A \cdot D}{B \cdot C}$.
  • Factor-First Strategy: Factorise all numerators and denominators prior to multiplication to permit cross-cancellation and prevent unnecessary polynomial expansion.

• Multiplication Problems:

  • Expression: $\frac{x}{4} \times \frac{12}{x^2}$
  • Multiply terms directly across: $\frac{12x}{4x^2}$.
  • Divide numerical coefficients by $4$: $\frac{12}{4} = 3$.
  • Divide variable factors using the exponent rule $\frac{x}{x^2} = \frac{1}{x}$: $\frac{3}{x}$.
  • Expression: $\frac{2x}{y^2} \times \frac{y^3}{6}$
  • Combine across the product: $\frac{2x \cdot y^3}{6y^2}$.
  • Simplify numerical coefficients: $\frac{2}{6} = \frac{1}{3}$.
  • Simplify power terms: $\frac{y^3}{y^2} = y$.
  • Resulting single simplified fraction: $\frac{xy}{3}$.
  • Expression: $\frac{x^2 - 9}{2x + 6} \times \frac{x^2 + 2x}{x^2 - 3x}$
  • Completely factorise each component:
    • $x^2 - 9 = (x - 3)(x + 3)$
    • $2x + 6 = 2(x + 3)$
    • $x^2 + 2x = x(x + 2)$
    • $x^2 - 3x = x(x - 3)$
  • Express as a combined product of factors: $\frac{(x - 3)(x + 3)}{2(x + 3)} \times \frac{x(x + 2)}{x(x - 3)}$.
  • Cancel identical terms $(x + 3)$, $(x - 3)$, and $x$ across numerator and denominator:       $\frac{1}{2} \times \frac{x + 2}{1} = \frac{x + 2}{2}$.

• Division Problems:

  • Expression: $\frac{x}{2} \div 4$
  • Express the integer $4$ as the rational fraction $\frac{4}{1}$.
  • Multiply by the reciprocal $\frac{1}{4}$: $\frac{x}{2} \times \frac{1}{4} = \frac{x}{8}$.
  • Expression: $\frac{x^2}{10} \div \frac{x}{5}$
  • Multiply by the reciprocal $\frac{5}{x}$: $\frac{x^2}{10} \times \frac{5}{x}$.
  • Multiply numerators and denominators: $\frac{5x^2}{10x}$.
  • Cancel the common factor of $5x$: $\frac{x}{2}$.
  • Expression: $\frac{2}{x^2 - 4} \div \frac{4}{x + 2}$
  • Invert the divisor to form a multiplication: $\frac{2}{x^2 - 4} \times \frac{x + 2}{4}$.
  • Factorise the difference of squares in the denominator: $\frac{2}{(x - 2)(x + 2)} \times \frac{x + 2}{4}$.
  • Cancel the common factor $(x + 2)$: $\frac{2}{4(x - 2)}$.
  • Simplify numerical factors $\frac{2}{4} = \frac{1}{2}$: $\frac{1}{2(x - 2)}$.

Addition and Subtraction of Algebraic Fractions

• Methods for Combining Rational Expressions:

  • Fractions cannot be combined additively without finding a Lowest Common Denominator (LCD).
  • Convert each fraction to an equivalent form possessing the LCD by multiplying the numerator and denominator by the missing factors.
  • When subtracting a fraction with a compound polynomial numerator, distribute the negative sign across every term in that numerator.

• Simplification to a Single Fraction with Numerical Denominators:

  • Expression: $\frac{x}{2} + \frac{x}{10}$
  • The Lowest Common Denominator of $2$ and $10$ is $10$.
  • Express with LCD: $\frac{5x}{10} + \frac{x}{10}$.
  • Combine numerators: $\frac{5x + x}{10} = \frac{6x}{10}$.
  • Reduce by dividing out the common factor of $2$: $\frac{3x}{5}$.
  • Expression: $\frac{x}{4} - \frac{x}{6}$
  • The Lowest Common Denominator of $4$ and $6$ is $12$.
  • Convert denominators: $\frac{3x}{12} - \frac{2x}{12}$.
  • Subtract numerators: $\frac{3x - 2x}{12} = \frac{x}{12}$.
  • Expression: $\frac{2x}{3} - \frac{x}{12}$
  • The Lowest Common Denominator of $3$ and $12$ is $12$.
  • Convert the first fraction: $\frac{4(2x)}{12} = \frac{8x}{12}$.
  • Subtract: $\frac{8x - x}{12} = \frac{7x}{12}$.
  • Expression: $\frac{x}{6} + \frac{2x}{3}$
  • The Lowest Common Denominator is $6$.
  • Convert the second fraction: $\frac{2(2x)}{6} = \frac{4x}{6}$.
  • Add: $\frac{x + 4x}{6} = \frac{5x}{6}$.
  • Expression: $1 - \frac{x}{7}$
  • Write the integer $1$ as $\frac{7}{7}$.
  • Combine over common denominator: $\frac{7 - x}{7}$.
  • Expression: $2 + \frac{x}{3} + \frac{x}{6}$
  • The Lowest Common Denominator of $1$, $3$, and $6$ is $6$.
  • Convert each term:
    • $2 = \frac{12}{6}$
    • $\frac{x}{3} = \frac{2x}{6}$
    • $\frac{x}{6} = \frac{x}{6}$
  • Combine numerators: $\frac{12 + 2x + x}{6} = \frac{12 + 3x}{6}$.
  • Factorise numerator: $\frac{3(4 + x)}{6}$.
  • Cancel factor of $3$: $\frac{x + 4}{2}$.

• Simplification with Compound and Variable Denominators:

  • Expression: $\frac{x - 1}{2} - \frac{2x - 3}{4}$
  • The common denominator is $4$.
  • Express the first fraction over $4$: $\frac{2(x - 1)}{4}$.
  • Combine under one fraction, placing the second numerator in parentheses to ensure correct sign distribution: $\frac{2(x - 1) - (2x - 3)}{4}$.
  • Expand the numerator: $\frac{2x - 2 - 2x + 3}{4}$.
  • Simplify like terms: $\frac{(2x - 2x) + (-2 + 3)}{4} = \frac{1}{4}$.
  • Expression: $\frac{1}{x + 1} - \frac{2}{x + 2}$
  • The algebraic LCD is the product of the coprime linear factors: $(x + 1)(x + 2)$.
  • Convert fractions: $\frac{1(x + 2)}{(x + 1)(x + 2)} - \frac{2(x + 1)}{(x + 1)(x + 2)}$.
  • Combine numerators: $\frac{(x + 2) - 2(x + 1)}{(x + 1)(x + 2)}$.
  • Expand and collect like terms: $\frac{x + 2 - 2x - 2}{(x + 1)(x + 2)} = \frac{-x}{(x + 1)(x + 2)}$.
  • Expression: $\frac{1}{x + 2} - \frac{2}{x} + \frac{1}{x - 2}$
  • The Lowest Common Denominator across all three denominators is $x(x + 2)(x - 2) = x(x^2 - 4)$.
  • Grouping symmetric terms first simplifies the arithmetic:       $\frac{1}{x - 2} + \frac{1}{x + 2} = \frac{(x + 2) + (x - 2)}{(x - 2)(x + 2)} = \frac{2x}{x^2 - 4}$.
  • Subtract the middle term $\frac{2}{x}$:       $\frac{2x}{x^2 - 4} - \frac{2}{x} = \frac{2x(x) - 2(x^2 - 4)}{x(x^2 - 4)}$.
  • Expand and simplify the numerator: $\frac{2x^2 - 2x^2 + 8}{x(x^2 - 4)} = \frac{8}{x(x^2 - 4)}$.

Number Theory and Parity Proofs with Rational Expressions

• Problem Statement:

  • Given that $m$ is an even integer and $m \neq -10$, prove that $\frac{70 - m^2 - 3m}{m + 10}$ is an odd integer.

• Algebraic Simplification:

  • Rearrange the quadratic numerator in standard descending order: $-m^2 - 3m + 70$.
  • Factor out $-1$: $-(m^2 + 3m - 70)$.
  • Find two numbers whose product is $-70$ and whose sum is $+3$: these values are $+10$ and $-7$.
  • Factor the quadratic expression: $-(m + 10)(m - 7) = (m + 10)(7 - m)$.
  • Substitute back into the rational expression:     $\frac{70 - m^2 - 3m}{m + 10} = \frac{(m + 10)(7 - m)}{m + 10}$.
  • Since $m \neq -10$, the factor $(m + 10) \neq 0$ and can be cancelled: $7 - m$.

• Parity Analysis:

  • By definition of an even integer, $m$ can be represented as $m = 2k$, where $k \in \mathbb{Z}$.
  • Substitute $m = 2k$ into the simplified expression:     $7 - m = 7 - 2k$.
  • Rewrite the expression in standard odd-integer form $2j + 1$:     $7 - 2k = (6 + 1) - 2k = 2(3 - k) + 1$.
  • Because $k$ is an integer, $j = 3 - k$ is also an integer.
  • Any integer of the form $2j + 1$ is strictly odd.
  • Additionally, the difference between an odd integer ($7$) and an even integer ($m$) is universally an odd integer ($\text{odd} - \text{even} = \text{odd}$).
  • Therefore, the expression always evaluates to an odd integer.

Solving Fractional and Rational Equations

• Systematic Procedure for Solving Rational Equations:

  • Identify the domain restrictions where denominators equal zero.
  • Clear all fractions by multiplying every term on both sides of the equation by the LCD of the denominators.
  • For single-fraction proportions of the form $\frac{A}{B} = \frac{C}{D}$, apply cross-multiplication: $A \cdot D = B \cdot C$.
  • Solve the resulting linear or quadratic polynomial equation.
  • Check solutions against domain restrictions to discard any extraneous solutions.

• Worked Equation Solutions:

  • Equation: $\frac{1 - 2x}{4} = \frac{x - 3}{2}$
  • Clear fractions by multiplying both sides by the LCD, $4$:       $1 - 2x = 4 \left(\frac{x - 3}{2}\right)$$1 - 2x = 2(x - 3)$
  • Expand the right side: $1 - 2x = 2x - 6$.
  • Collect variable terms on one side and constants on the other:       $1 + 6 = 2x + 2x$$7 = 4x$$x = \frac{7}{4}$.
  • Equation: $\frac{2x + 1}{3} - \frac{x - 2}{4} = -2$
  • Clear denominators by multiplying the entire equation by $12$:       $12 \left(\frac{2x + 1}{3}\right) - 12 \left(\frac{x - 2}{4}\right) = 12(-2)$$4(2x + 1) - 3(x - 2) = -24$
  • Expand brackets:       $8x + 4 - 3x + 6 = -24$
  • Combine like terms:       $5x + 10 = -24$$5x = -34$$x = -\frac{34}{5} = -6.8$.
  • Equation: $\frac{10}{x + 2} = 2$
  • Multiply both sides by the denominator $(x + 2)$, given restriction $x \neq -2$:       $10 = 2(x + 2)$$10 = 2x + 4$$6 = 2x$$x = 3$.

• Equation: $\frac{2}{x + 3} = \frac{3}{x - 1}$

  • Apply cross-multiplication, noting domain restrictions $x \neq -3$ and $x \neq 1$:       $2(x - 1) = 3(x + 3)$
  • Expand:     $2x - 2 = 3x + 9$
  • Isolate $x$:     $-2 - 9 = 3x - 2x$$x = -11$.

• Equation: $\frac{x + 2}{x - 2} = \frac{5}{3}$

  • Cross-multiply with restriction $x \neq 2$:       $3(x + 2) = 5(x - 2)$
  • Expand brackets:     $3x + 6 = 5x - 10$
  • Rearrange terms:     $6 + 10 = 5x - 3x$$16 = 2x$$x = 8$.

• Equation: $\frac{x}{x - 2} + \frac{x}{x + 1} = 2$

  • Domain restrictions: $x \neq 2$ and $x \neq -1$.

• Multiply through by the LCD, $(x - 2)(x + 1)$:   $x(x + 1) + x(x - 2) = 2(x - 2)(x + 1)$

• Expand all terms:   $(x^2 + x) + (x^2 - 2x) = 2(x^2 - x - 2)$$2x^2 - x = 2x^2 - 2x - 4$

• Subtract $2x^2$ from both sides to reduce the quadratic to a linear equation:   $-x = -2x - 4$

• Add $2x$ to both sides:   $x = -4$.

• Check restriction: $x = -4$ does not violate $x \neq 2$ or $x \neq -1$, confirming the solution.

Mixture and Concentration Problems

• Mathematical Framework of Dilution:

  • Concentration formula: $\text{Concentration} = \frac{\text{Quantity of Pure Solute}}{\text{Total Volume of Solution}}$.
  • Adding pure water ($0\%$ concentration) increases the total liquid volume of the mixture while keeping the quantity of active solute strictly constant.

• Problem Statement:

  • How much water must be added to $250\,\text{mL}$ of a $35\%$ cordial mixture so that the resulting mixture is $10\%$ cordial?

• Step-by-Step Calculation:

  • Determine Initial Solute (Pure Cordial) Volume:$\text{Volume of pure cordial} = 35\% \times 250\,\text{mL} = 0.35 \times 250 = 87.5\,\text{mL}$.
  • Define Unknowns:
  • Let $w$ represent the volume of water (in $\text{mL}$) to be added.
  • New total volume of mixture: $250 + w\,\text{mL}$.
  • Set up the Final Concentration Equation:
  • The desired concentration is $10\% = 0.10 = \frac{1}{10}$.
  • $\frac{87.5}{250 + w} = 0.10$.
  • Solve for Added Water ($w$):$87.5 = 0.10(250 + w)$$87.5 = 25 + 0.10w$$87.5 - 25 = 0.10w$$62.5 = 0.10w$$w = \frac{62.5}{0.10} = 625\,\text{mL}$.
  • Conclusion:
  • Exactly $625\,\text{mL}$ of water must be added to dilute the solution to a $10\%$ cordial concentration.

Comparative Fractions, Sequences, and Algebraic Proofs

• Numerical Comparison of Unit Fraction Pairs:

  • Pair 1: Determine which is larger: $\frac{1}{2} + \frac{1}{4}$ or $\frac{1}{3} + \frac{1}{3}$.
  • First sum: $\frac{1}{2} + \frac{1}{4} = \frac{2}{4} + \frac{1}{4} = \frac{3}{4} = 0.75$.
  • Second sum: $\frac{1}{3} + \frac{1}{3} = \frac{2}{3} \approx 0.6667$.
  • Comparison using common denominator $12$: $\frac{9}{12} > \frac{8}{12}$.
  • Result: $\frac{1}{2} + \frac{1}{4}$ is larger.
  • Pair 2: Determine which is larger: $\frac{1}{4} + \frac{1}{6}$ or $\frac{1}{5} + \frac{1}{5}$.
  • First sum: $\frac{1}{4} + \frac{1}{6} = \frac{3}{12} + \frac{2}{12} = \frac{5}{12} = \frac{25}{60} \approx 0.4167$.
  • Second sum: $\frac{1}{5} + \frac{1}{5} = \frac{2}{5} = \frac{24}{60} = 0.40$.
  • Comparison: $\frac{25}{60} > \frac{24}{60}$.
  • Result: $\frac{1}{4} + \frac{1}{6}$ is larger.
  • Pair 3: Determine which is larger: $\frac{1}{8} + \frac{1}{10}$ or $\frac{1}{9} + \frac{1}{9}$.
  • First sum: $\frac{1}{8} + \frac{1}{10} = \frac{5 + 4}{40} = \frac{9}{40} = \frac{81}{360} = 0.225$.
  • Second sum: $\frac{1}{9} + \frac{1}{9} = \frac{2}{9} = \frac{80}{360} \approx 0.2222$.
  • Comparison: $\frac{81}{360} > \frac{80}{360}$.
  • Result: $\frac{1}{8} + \frac{1}{10}$ is larger.

• Generalisation for Positive Integers ($x > 1$):

  • Comparing $\frac{1}{x - 1} + \frac{1}{x + 1}$ with $\frac{1}{x} + \frac{1}{x} = \frac{2}{x}$.
  • Following the inductive pattern observed in numerical cases ($x = 3$, $x = 5$, $x = 9$):     $\frac{1}{x - 1} + \frac{1}{x + 1} > \frac{1}{x} + \frac{1}{x}$.
  • The asymmetric spread pair centered at $x$ is always strictly greater than the symmetric pair at $x$.

• Formal Proof via Algebraic Simplification:

  • Consider the difference: $\Delta = \frac{1}{x - 1} + \frac{1}{x + 1} - \frac{2}{x}$.
  • Combine the first two terms over their common denominator $(x - 1)(x + 1) = x^2 - 1$:     $\frac{1}{x - 1} + \frac{1}{x + 1} = \frac{(x + 1) + (x - 1)}{(x - 1)(x + 1)} = \frac{2x}{x^2 - 1}$.
  • Subtract $\frac{2}{x}$ by finding a common denominator $x(x^2 - 1)$:     $\Delta = \frac{2x}{x^2 - 1} - \frac{2}{x} = \frac{2x(x) - 2(x^2 - 1)}{x(x^2 - 1)}$.
  • Expand and simplify the numerator:     $\Delta = \frac{2x^2 - (2x^2 - 2)}{x(x^2 - 1)} = \frac{2x^2 - 2x^2 + 2}{x(x^2 - 1)} = \frac{2}{x(x^2 - 1)}$.
  • Analyze the sign of $\Delta$ for any integer $x > 1$:
  • The numerator is $2$, which is strictly positive ($2 > 0$).
  • Because $x > 1$, the factor $x$ is strictly positive ($x > 0$).
  • Because $x > 1$, $x^2 > 1$, which ensures $(x^2 - 1) > 0$.
  • The product of positive terms in the denominator is strictly positive: $x(x^2 - 1) > 0$.
  • Consequently, the entire quotient is strictly positive: $\frac{2}{x(x^2 - 1)} > 0$.
  • Because $\left(\frac{1}{x - 1} + \frac{1}{x + 1}\right) - \frac{2}{x} > 0$, it is rigorously proven that:     $\frac{1}{x - 1} + \frac{1}{x + 1} > \frac{1}{x} + \frac{1}{x}$     for all integers $x > 1$.